So we have an the equation $\frac{2}{3}t^2+\frac{4}{3}t=\frac 15$, when you finish solving the equation you get $t = \frac{-10 + \sqrt{130}}{10} $ and $\frac{-10 - \sqrt{130}}{10}$.
The text book seems to think the solution it $\frac{10 + \sqrt{130}}{10}$ and $\frac{10 - \sqrt{130}}{10}$.
I've tried checking other sources to verify my answer and just want to be doubly sure that I have the right solution, so any input would be appreciated.
Multiplying the both sides of $$\frac{2}{3}t^2+\frac 43t=\frac 15$$ by $15$ gives us $$10t^2+20t=3\Rightarrow 10t^2+20t-3=0\Rightarrow t=\frac{-10\pm\sqrt{130}}{10}.$$
Here, note that $$ax^2+\color{red}{2}bx+c=0\Rightarrow x=\frac{-b\pm\sqrt{b^2-ac}}{a}.$$